Sure, I'd be happy to help you solve this system of equations!
First, let's solve the first equation for x:
x = 1 - 2y
Now, substitute this expression for x into the second equation:
y² - 2y - 3(1 - 2y) + 1 - 2(1 - 2y)² = 3(1 - 2y)
Expand and simplify:
y² - 2y - 3 + 1 - 4y + 4y² = 3
Combine like terms:
y² - 2y - 3 + 4y² = 7
Simplify:
y² - 2y - 3 = 7
Now, we can factor the left-hand side:
(y - 1)(y + 3) = 7
So, we have two possible solutions:
y - 1 = 0 or y + 3 = 0
Solving for y, we get:
y = 1 or y = -3
Now, let's check our solutions. Substituting y = 1 into the first equation, we get:
x = 1 - 2(1) = 0
This is a valid solution, so (x, y) = (0, 1) is a solution to the system.
Substituting y = -3 into the first equation, we get:
x = 1 - 2(-3) = 7
This is also a valid solution, so (x, y) = (7, -3) is another solution to the system.
Therefore, the solutions to the system are (x, y) = (0, 1) and (x, y) = (7, -3).